TITLE: The value of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log n}{n}$
QUESTION [5 upvotes]: How to compute the following convergent series? or some hint!
$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log n}{n}.$$
The Wolfram MATHEMATICA9.0 gives the result is $1/2(\log 2)^2-\gamma\log 2$,
where $\gamma$ is tha Euler constant!

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n}:\
{\large ?}}$.

Hereafter, I'll use
well known identities which are related to the Riemann Zeta Function:
\begin{align}
&\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n} =
\left.\partiald{}{\mu}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1 - \mu}}
\right\vert_{\ \mu\ =\ 0^{+}} =
\left.\partiald{\bracks{\pars{1 - 2^{\mu}}\zeta\pars{1 - \mu}}}{\mu}\,
\right\vert_{\ \mu\ =\ 0^{+}}
\\[5mm] = &\
\partiald{}{\mu}
\braces{\bracks{-\ln\pars{2}\mu - {1 \over 2}\,\ln^{2}\pars{2}\,\mu^{2} +
\,\mrm{O}\pars{\mu^{3}}}\bracks{-\,{1 \over \mu} + \gamma + \,\mrm{O}\pars{\mu^{1}}}}_{\ \mu\ =\ 0^{+}}
\\[5mm] = &\
\partiald{}{\mu}\braces{\ln\pars{2} +
\bracks{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}}\mu + \,\mrm{O}\pars{\mu^{2}}}_{\ \mu\ =\ 0^{+}}
\\[5mm] = &\
\bbx{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}}
\end{align}